TITLE: Showing $ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}}$
QUESTION [21 upvotes]: I would like to show that:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} $$
We have:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\sum_{n=0}^{\infty} \frac{1}{3n+1}-\frac{1}{3n+2} $$
I wanted to use the fact that $$\arctan(\sqrt{3})=\frac{\pi}{3} $$ but $\arctan(x)$ can only be written as a power series when $ -1\leq x \leq1$...

REPLY [4 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\sum_{n=0}^{\infty}\frac{1}{\pars{3n + 1}\pars{3n + 2}}}
=\frac{1}{9}\sum_{n=0}^{\infty}\frac{1}{\pars{n + 1/3}\pars{n + 2/3}}
=\frac{1}{3}\bracks{\Psi\pars{\frac{2}{3}} - \Psi\pars{\frac{1}{3}}}
\end{align}
where $\ds{\Psi}$ is the Digamma Function.
With Euler Reflection Formula:
\begin{align}&\color{#66f}{\large%
\sum_{n=0}^{\infty}\frac{1}{\pars{3n + 1}\pars{3n + 2}}}
=\frac{1}{3}\bracks{\pi\cot\pars{\pi\,\frac{1}{3}}}
=\color{#66f}{\large\frac{\pi}{3\root{3}}}
\end{align}
because $\ds{\cot\pars{\frac{\pi}{3}} = {1 \over \root{3}}}$.